Two light bulbs, one 30 ohm and the other 60 ohm, are connected in parallel to a

Question

Two light bulbs, one 30 ohm and the other 60 ohm, are connected in parallel to a 120 Volt battery (all three circuit elements are in parallel). In this situation: the current in the 60 ohm bulb is less than that in the 30 ohm bulb. the 60 ohm bulb will be brighter than the 30 ohm bulb. both bulbs will light with equal brightness. each bulb will have a potential difference of 60 Volts. the current in the 60 ohm bulb is equal to that in the 30 ohm bulb. In the picture below, a current of 5.0 A goes through the resistor labeled R and a current of 1.0 A goes through the 3.0 ohm resistor. What is voltage, V, of the battery? What is the equivalent resistance for these resistors between points a and b? A circuit contains a voltage source (of unknown voltage, DeltaV) and three identical resistors, each of resistance R. The current going through the rightmost resistor (I1) is 2.0 Amps. What is the magnitude of the current that leaves the voltage source, I3?

Explanation / Answer

1) a is the correct options since current in 30 ohm=120/30

=4 A
current n 60=120/60

=2 A

power= V^2/r.

and since in parallel, both will have 120 V p.d.

2) R*5=1*3

or R=0.6 ohm

so current throug the cell=5+1

=6 A

net resistance=4 + (0.6*3)/(0.6+3)

=4.5 ohm

so V=4.5*6

=27 V (a)

3)Req=7+1+3+ 8*4/(8+4)

=13.7 ohm (e)

4)current =2+1

=3 A (c)

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